I was reading an article about Einstein-Cartan universes and the beginning he defines the contortion tensor as $$ K^a\vphantom{K}_{bc}=S^a\vphantom{S}_{bc}+S_{bc}\vphantom{S}^a+S_{cb}\vphantom{S}^a $$ with $$S^a\vphantom{S}_{bc}=\Gamma^a\vphantom{S}_{[bc]}.$$

Then he writes two equalities that I cannot understand:

$$K_{(a|b|c)}=-2S_{(ac)b}$$ and $${ }K_{[a|b|c]}=-S_{bac}. $$

What are the definition of $K_{(a|b|c)}$ and ${ }K_{[a|b|c]}$?


Obviously, $K_{abc} = g_{ad}{K^d}_{bc}$. And I'm guessing you know that $()$ means symmetrization and $[]$ means antisymmetrization over whatever indices are contained between them.

The vertical bars are there to "suspend" the symmetrization or anti-symmetrization that is happening on the other indices. That is, the $b$ in each case is not involved: \begin{gather} K_{(a|b|c)} = \frac{1}{2} \left( K_{abc} + K_{cba} \right), \\ K_{[a|b|c]} = \frac{1}{2} \left( K_{abc} - K_{cba} \right). \end{gather}

So, for example, it's straightforward to compute \begin{align} K_{(a|b|c)} &= \frac{1}{2} \left( K_{abc} + K_{cba} \right) \\ &= \frac{1}{2} \left( S_{abc} + S_{bca} + S_{cba} + S_{cba} + S_{bac} + S_{abc} \right) \\ &= S_{abc} + S_{b(ca)} + S_{cba} \\ &= S_{abc} + S_{cba} \\ &= -(S_{acb} + S_{cab}) \\ &= -2S_{(ac)b} \end{align} In the third line, I eliminated $S_{b(ca)}$ because it is totally antisymmetric on the last two indices (by definition), so the symmetric part of those two indices is zero. In the fourth line, I swapped the last two indices, and flipped the sign, because of that same antisymmetry.

Similarly, we have \begin{align} K_{[a|b|c]} &= \frac{1}{2} \left( K_{abc} - K_{cba} \right) \\ &= \frac{1}{2} \left( S_{abc} + S_{bca} + S_{cba} - S_{cba} - S_{bac} - S_{abc} \right) \\ &= \frac{1}{2} \left( S_{bca} - S_{bac} \right) \\ &= S_{b[ca]} \\ &= -S_{b[ac]} \\ &= -S_{bac} \end{align} All of these manipulations should become very obvious once you get a little practice with index gymnastics.

  • $\begingroup$ Computing the second equation I found out that is equal to $1/2(S_{bca}-S_{bac})$ that differs from $-S_{bac}$. Where am I wrong? $\endgroup$ – raskolnikov Mar 16 '17 at 14:15
  • $\begingroup$ You did the computation right, you just need to go one step further. What you got is the same as $-S_{bac}$. Look at the definition of $S$: it's antisymmetric under exchange of its last two indices. Thus, your first term $S_{bca}$ is equal to $-S_{bac}$. $\endgroup$ – Mike Mar 16 '17 at 14:24
  • 1
    $\begingroup$ Please be careful: index gymnastics with the contorsion tensor may seriously harm your fingers :-) $\endgroup$ – magma Mar 18 '17 at 1:54

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